CHM 211
Fall 2015
Test 1
Name: ____________________
(Please print)
Multiple Choice: (4 points each. Put answers in left margin as capital letters.)
1. A block has a temperature of 185 K. What is its temperature in ºC?
A) 167
B) 185
C) 366
D) 458
15 x 40.0  13.5
?
5.000 x 103
2. How many significant figures are in the answer to the problem:
A) 1
B) 2
C) 3
E) 585
D) 4
E) 5
3. Which of the following is not an exact number?
A) density of pure iron-56
B) mass of one carbon-12 atom
C) one dozen
D) protons in a carbon nucleus
E) seconds in a minute
4. Which of the following is not a unit typically used in chemistry?
A) ºC
B) g
C) in
D) K
E) L
5. Which of the following properties is usually associated with metals?
A) brittle
B) dull
C) insulators
D) malleable
E) poor heat conduction
6. The correct number of protons, neutrons, and electrons in I- (I-127) is:
A) 53p, 74n, 52e
B) 53p, 74n, 54e
C) 54p, 73n, 53e
D) 74p, 53n, 75e
E) 76p, 52n, 78e
7. Which of the following is least likely to represent a real compound?
A) Al2(CO3)3
B) BF2
C) NaBr
D) O3
E) TiO2
8. Which of the following elements reacts most similarly to phosphorus?
A) C
B) Ca
C) F
D) N
E) S
9. Which of the following is an example of a combustion reaction?
A) N2 + 3 H2  2 NH3
B) 4 Fe + 3 O2  2 Fe2O3
C) AgNO3 + NaCl  AgCl + NaNO3
D) C3H8 + 5 O2  3 CO2 + 4 H2O
E) CaCO3  CaO + CO2
Discussion Questions: (You must show your work to receive credit.)
1. Define: (12 points)
accuracy – a measure of how close a measured value is to the “real” or “true” value
cation – positively charged ions
molecular formula – a chemical formula with the exact number of each kind of atom
2. Give one example of each of the following: (6 points)
halogen: F, Cl, Br, I, At
alkaline earth metal: Be, Mg, Ca, Sr, Ba, Ra
chalcogen: O, S, Se, Te, Po
3. Complete the following: (11 points)
BCl3(g) + 3 H2O()  H3BO3 (s) + 3 HCl(g)
When solutions of calcium chloride and sodium phosphate are mixed, solid calcium
phosphate forms and sodium chloride remains in solution.
3 CaCl2 (aq) + 2 Na3PO4 (aq)  Ca3(PO4)2 (s) + 6 NaCl(aq)
4. For the following, give the name or formula where appropriate: (15 points)
CuCl2 – copper(II) chloride
potassium bromide – KBr
NH4(C2H3O2) – ammonium acetate
sodium nitride – Na3N
sulfuric acid – H2SO4
5. Sodium hydroxide reacts with carbon dioxide according to the reaction below. If 5.21 g of
carbon dioxide are bubbled through a solution containing 4.38 g of sodium hydroxide, how
much sodium hydrogen carbonate will be produced? How much of which reagent will be left
over? (10 points)
NaOH(aq) + CO2 (g)  NaHCO3 (aq)
Determine the limiting reagent by first calculating the number of moles of each substance
present.
1 mol
molCO2 = (5.21 g)(44.01 gCO2 ) = 0.118 molCO2
CO2
1 mol
molNaOH = (4.38 g)(40.00 gNaOH ) = 0.110 molNaOH
NaOH
Now determine how much carbon dioxide is required to use up all of the sodium hydroxide.
1 mol
molCO2(required) = (0.110 molNaOH) (1 mol CO2 ) = 0.110 molCO2
NaOH
But you have 0.118 molCO2 to begin with, which is more than enough carbon dioxide so all
of the sodium hydroxide is consumed. Thus, sodium hydroxide is the limiting reagent.
1 𝑚𝑜𝑙
1 𝑚𝑜𝑙
84.01 𝑔
massNaHCO3 = (4.38 gNaOH)(40.00 𝑔𝑁𝑎𝑂𝐻 )( 1 𝑚𝑜𝑙𝑁𝑎𝐻𝐶𝑂3 )( 1 𝑚𝑜𝑙 𝑁𝑎𝐻𝐶𝑂3 ) = 9.20 gNaHCO3
𝑁𝑎𝑂𝐻
𝑁𝑎𝑂𝐻
𝑁𝑎𝐻𝐶𝑂3
massCO2(excess) = 5.21 g + 4.38 g – 9.20 g = 0.39 gCO2
6. Almost all chromium is smelted from chromite ore, whose composition is 46.5% chromium,
25.0% iron, and 28.6% oxygen. The mineral weighs about 225 g per mole. What are its
empirical and molecular formulas? (10 points)
Assume 100 g of compound.
1 mol
molCr = (46.5 gC(51.996 Crg )r) = 0.894 molCr
(0.448 molCr ) = 2.00 molCr
molFe = (25.0 gFe)(55.845 g ) = 0.448 molFe
(0.448 mol ) = 1
0.894 mol
Cr
1 molFe
1 molO

Fe
Fe
O
eu
1 mol
)(51.996 Crg ) + (
Cr
1 molFe
eu
Fe
1 mol
)(55.845 Fe
)+(
g
Fe
4 molOr
eu
= 223.842 g/eu [eu = empirical unit]
223.842 g
1 mol
formula units = ( eu )( 225 g)= 0.995 mol/eu ≈ 1 mol/eu
molecular formula is Cr2FeO4
mol
(0.448 mol ) = 4.00 mol O
 empirical formula is Cr2FeO4
2 molCr
Fe
1.79 molO
molO = (28.6 gO) (15.9994 g ) = 1.79 molO
EWCr2FeO4 = (
mol
Fe
0.448 molFe
Fe
1 mol
)(15.9994Og )
O